Answer
See below
Work Step by Step
Given: $y''-2y'+10y=\cos 2t$
Since $y(0)=0\\
y'(0)=1$
obtain: $L[y]=\frac{1}{s^2-2s+10}+\frac{s}{(s^2+4)(s^2-2s+10)}$
Using the Convolution Theorem
$y(t)=L^{-1}[\frac{1}{s^2-2s+10}]+L^{-1}[\frac{s}{(s^2+4)(s^2-2s+10)}]\\
=\frac{1}{3}e^t\sin 3t+\int^t_0 e^{\tau} \sin (3\tau) \cos [2(t-\tau)]d \tau $
